Define $f(x, y, z) = x + y + z$. Let $\vec{a} = (1, 2, 1)$ and $\vec{v} = \left( -\dfrac{\sqrt{2}}{2}, 0, \dfrac{\sqrt{2}}{2} \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Solution: Our ultimate goal is to substitute $h = 0$ into the limit and get the answer. To do this, we first need to cancel out the $h$ in the denominator. Otherwise we're dividing by zero. How can we simplify the limit so that we get an $h$ on the top and an $h$ on the bottom, which we can then cancel? Let's plug in $\vec{a}$ and $\vec{v}$ to the limit. $ \lim_{h \to 0} \dfrac{f \left( (1, 2, 1) + h \left( -\dfrac{\sqrt{2}}{2}, 0, \dfrac{\sqrt{2}}{2} \right) \right) - f(1, 2, 1)}{h}$ Now we can add together the vectors. $ \lim_{h \to 0} \dfrac{f \left(1 - \dfrac{\sqrt{2}}{2} h, 2, 1 + \dfrac{\sqrt{2}}{2} h \right) - f(1, 2, 1)}{h}$ Let's evaluate $f$. $ \lim_{h \to 0} \dfrac{1 - \dfrac{\sqrt{2}}{2} h + 2 + 1 + \dfrac{\sqrt{2}}{2} h - (1 + 2 + 1)}{h}$ We can cancel out all the terms without an $h$. $ \lim_{h \to 0} \dfrac{- \dfrac{\sqrt{2}}{2} h + \dfrac{\sqrt{2}}{2} h}{h}$ Now we can cancel $h$ and calculate the limit. $\begin{aligned} \lim_{h \to 0} \dfrac{- \dfrac{\sqrt{2}}{2} h + \dfrac{\sqrt{2}}{2} h}{h} &= \lim_{h \to 0} 0 \\ \\ &= 0 \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = 0$.